If the energy of the $n^{th}$ orbit in a hydrogen atom is $E_n$,what will be the energy of the $n^{th}$ orbit in a helium ion $(He^+)$?

In the hydrogen atom, the electron makes a transition from the higher orbit $(i)$ to a lower orbit $(f)$. The ratio of the radius of the orbits is given by $r_i : r_f = 16 : 4$. The wavelength of the photon emitted due to this transition is . . . . . . nm. (Given Rydberg constant $R = 1.0973 \times 10^7 \text{ m}^{-1}$)

The figure shows a graph between $\ln \left| \frac{A_n}{A_1} \right|$ and $\ln |n|$, where $A_n$ is the area enclosed by the $n^{th}$ orbit in a hydrogen-like atom. The correct curve is

The ground state energy of a hydrogen atom is $-13.6 \; eV$. What are the kinetic and potential energies of the electron in this state?

The radius of the first orbit of hydrogen is $r_{H}$,and the energy in the ground state is $-13.6 \text{ eV}$. Considering a $\mu^{-}$-particle with a mass $207 m_e$ revolving around a proton as in a hydrogen atom,the energy and radius of the proton and $\mu^{-}$-combination respectively in the first orbit are (assume the nucleus to be stationary):

Monochromatic radiation of wavelength $\lambda$ is incident on a hydrogen sample containing atoms in the ground state. Hydrogen atoms absorb the light and subsequently emit radiations of ten different wavelengths. The value of $\lambda$ is.....$nm$.